In this section, we show that in the case of three correlated bidders, it is NP-complete to decide whether there exists a truthful auction such that the probability of getting revenueT is at leastQfor the given distributionDand target revenueT. Specifically, the input of the problem is the distribution D expressed explicitly as its probability mass function in the size ofD, the target revenueT and a probabilityQ. We state our main theorem in this section.
Theorem 4.9. It is NP-complete to compute an optimal auction for three correlated bidders.
i i−k n−i n−i+k n−i+k+t t k k i=n n−k 0 k k+t t k n n−k=t 0 k t k n i i−k n−i n−i+k n t k k bidder 1 value bidder 2 value bidder 1 value bidder 2 value bidder 1 value bidder 2 value bidder 1 value bidder 2 value
Figure 4.6: Illustration of the dynamic programming.
Proof. Membership in NP follows from noting that any truthful auction A can be expressed by three pricing functionsf1, f2, f3, that use space polynomial in the input size. So we can guess the values off1(v2, v3),f2(v1, v3) andf3(v1, v3) for all (v1, v2, v3)∈ V and then compute the probability of getting revenue T by considering the profiles one by one.
We show NP-hardness by reducing from the problem Vertex Cover. Given a
graph withGwithnvertices labeled{1,2, . . . , n}andmedges, we construct an instance with three bidders where the support of the valuation distributions of bidder 1 and bidder 2 are [n] and bidder 3’s valuation are taken from [n]∪ {1.5}1. We also set the target revenue T to be n+ 2. Then NP-hardness will be shown in two steps. First we show NP-hardness for finding an optimal auction for bidders 1 and 2, where bidder 3 has his valuation fixed at v3 = n, but contributes revenue according to a pricing functionh(v1, v2), where h is part of the input. Next we construct a distribution for v3 < nthat endogenizeshin the sense that any truthful auction with good performance must use h as the pricing function for bidder 3. First of all, we define the particular functionhas follows. h(v1, v2) = 0 ifv1+v2≤n 1 ifv1+v2=n+ 1 v1+v2−n−1 ifv1+v2> n+ 1
Now we consider the profiles withv3 =n. Sinceh(v1, v2)< n, the revenue from bidder
1
1 n 1 1 n n n 1 1 n n n 1 2 3 6 1 4 5 bidder 1 value bidder 2 value
Figure 4.7: An example for the construction of the distribution forv3=nwhen n= 6. The numbers 1 andnin the (v1, v2)-pair boxes are proportional to their probabilities. The corresponding auction is shown on the left if the vertex cover of the graph is
{1,4,5}shown as gray nodes on the right.
3 is exactly given byh, i.e. R3(v1, v2, n) =h(v1, v2) for allv1,v2. So the revenue needed from bidder 1 and 2 to achieve the target revenuen+ 2 isn+ 2−h(v1, v2). Recall that Pr[v] denotes the probability when the profile is v. For simplicity of presentation, we work with probabilities higher than 1 that can be normalized by dividing by the sum of all probabilities later. Given the graphG, we define a correlated distribution as follows (illustrated in Figure 4.7). Pr[(v1, v2, n)] = 0 if v1+v2≤n 1 if v1+v2=n+ 1 n ifv1+v2> n+ 1 and (v1, n+ 1−v2)∈E 0 if v1+v2> n+ 1 and (v1, n+ 1−v2)6∈E
We will show that Ghas a vertex cover with size at most k < n iff there exists an auction whose probability to achieve the target revenue is at leastm·n+n−kwherem is the number of edges inG. For necessity, supposeSis a vertex cover forGand|S| ≤k. We construct an auction such thatf1(v2, n) =n+ 2−v2 if n+ 1−v2 ∈ S otherwise f1(v2, n) =n+1−v2andf2(v1, n) =n+2−v1ifv1∈Sotherwisef2(v1, n) =n+1−v1. Moreover we set the pricing function f3 for bidder 3 to be the particular h defined above. Then for any (v1, v2) such thatv1+v2> n+ 1 and Pr[(v1, v2, n)]>0, we have (v1, n+ 1−v2)∈E by the construction of Pr[(v1, v2, n)]. Since S is a vertex cover for G, we have v1 ∈ S or n+ 1−v2 ∈ S. So the revenue from the profile (v1, v2, n) is f1(v1, n)+f2(v2, n)+f3(v1, v2) that is at leastn+1−v2+n+1−v1+v1+v2−n−1+1 = n+ 2. Thus the probability that these profiles contribute to the event of raising the target revenue is m·n. On the other hand, for any (v1, v2) such that v1+v2 =n+ 1 andv1 6∈S, the revenue of the profile (v1, v2, n) isn+ 1−v1+n+ 1−v2+ 1 =n+ 2. Therefore the total probability is at leastm·n+n−k.
For the sufficiency, supposeAis an auction whose probability is at leastm·n+n−k. Note that we assume Ause h as the pricing function for bidder 3. LetS be the set of verticesv1 such thatA doesn’t get the target revenue from the profile (v1, n+ 1−v1).
Since the probability achieved byAis at leastm·n+n−kandk < n, it is easy to see that
|S| ≤k and A achieves the target revenue for all profile (v1, v2) with Pr[v1, v2, n] =n. For any edge (v1, n+ 1−v2)∈E, it is w.l.o.g. to assumev1 > n+ 1−v2, i.e.,v1+v2> n+ 1. Then we have Pr[v1, v2, n] = n by the construction. So the revenue obtained from this profile byA is at leastn+ 2, i.e. f1(v2, n) +f2(v1, n) +f3(v1, v2)≥n+ 2. So we havef1(v2, n)> n+ 1−v2 orf2(v1, n)> n+ 1−v1. That isv1 ∈S orn+ 1−v2 ∈S. ThereforeS is a vertex cover of G.
Next, we show how to construct the distribution for v3 < nsuch that any auction with good performance must useh be the pricing function for bidder 3. We divide all profiles inton+ 1 layers according to the value ofv3. In layer 0,v3 = 1; in layer 1,v3= 1.5; and for allk= 2, . . . , n,v3 =k. The general idea is to use layerk∈ {0, . . . , n−1}
to makef3 equal toh on the profiles (v1, v2) such thatv1+v2=n+ 1 +k. Finally, we use the last layern(whenv3 =n) to encode thevertex cover instance as described above.
We present the inductive construction of the layers from 1 to n−1. For layer 0, only the profiles on the diagonal have positive probabilities, i.e. Pr[(v1, v2,1)] =n iff v1 +v2 = n+ 1. So in order to get all the probabilities in this layer, the auction must set the pricing function for bidder 3 as f3(v1, v2) = 1 for all (v1, v2) such that v1+v2 =n+ 1.
For layer 1, we first set Pr[(v1, n+ 2−v1,1.5)] =nfor allv1 ∈[n] and Pr[(v1, n+ 1−
v1,1.5)] =n for all oddv1. We will show in order to get all the positive probabilities from layer 0 and layer 1, the auction must setf3(v1, v2) = 1 for allv1+v2=n+2. Note that for any (v1, v2) such that v1+v2 =n+ 2, eitherv1 is odd orn+ 1−v2 =v1−1 is odd. That is either Pr[(v1, v2 −1,1.5)] = n or Pr[(v1 −1, v2,1.5)] = n by the construction. Sincef3(v1, v2−1) = f3(v1−1, v2) = 1 by the construction of layer 0, we have eitherf2(v1, n) =v2−1 or f1(v2, n) =v1−1 to guarantee the auction get all probabilities in layer 1. That isf1(v2, n) +f2(v1, n) =v1+v2−1 =n+ 1. So we need to setf3(v1, v2) ≥1 to get the target revenue. Since v3 = 1.5 in this layer, the value of f3(v1, v2) cannot be higher than 1.5. By Proposition 4.1, it is w.l.o.g. to set it 1 because the target revenue and the valuations of the other bidders are integers.
Finally, for layer k ≥ 2, we set Pr[(v1, v2,1.5)] = n for all v1 +v2 = n+ 1 +k. Then We construct a graphG0 withnvertices labeled by [n] and a set of edges denoted by E such that (v1, v2) ∈ E iff v1 +n+ 1−v2 = n+ 1 +k as shown in Figure 4.8. Since the graph consists ofkchains, there exists an exact coverS0 for this graph. That is for each (v1, v2) ∈ E, one and only one of {v1, v2} is in S0. Actually this can be proven constructively by settingS0 ={1,2, . . . , k,2k+ 1,2k+ 2, . . . ,3k, . . .}as labeled in Figure 4.8. Given this exact coverS0, we define the distribution for layerkas follows: Pr[(v1, n+ 1−v1, k)] =nfor allv1 ∈S0 and Pr[(v1, n+ 2−v1, k)] =nifn+ 1−v26∈S0 or v1 6∈ S0 as shown in Figure 4.8. Intuitively, we construct the distribution such
n n n n n n n 1 k+ 1 n−k+ 1 2 k+ 2 n−k+ 2 k 2k n n k+ 1 k+ 2 2k player 1 value pla yer 2 value n n n n k
Figure 4.8: An illustration for the construction of layerk.
The gray vertices form the exact coverS when n is a multiple of k and n/k is even. The left graph shows the construction and optimal auction whenn= 8 and k= 4.
that for any (v1, v2) with v1+v2 = n+ 1 +k, either Pr[(v1, n+ 1−v1, k)] = n and Pr[(n+2−v2, v2, k)] =nor Pr[(v1, n+2−v1, k)] =nand Pr[(n+1−v2, v2, k)] =n. These profiles will restrict the values off1(v2, k) andf2(v1, k). Furthermore, we require for any (v1, n+ 2−v1) with positive probability, not both (v1, n+ 1−v1) and (v1−1, n+ 2−v1) have positive probabilities to guarantee thatf2(v1, k) +f1(n+ 2−v1, k)≥n+ 1.
Now we prove that in order to get all the probabilities in first n layers from layer 0 to layer n−1, the auction must set h to be the pricing function for bidder 3. Let A= (f1, f2, f3) be the auction that gets the target revenue for all profiles with positive probabilities. It suffices to showf3(v1, v2) =h(v1, v2) for all (v1, v2) such thatv1+v2≥ n+ 1 since no profile with v1 +v2 < n+ 1 has positive probability. Recall that we have shown f3(v1, v2) = h(v1, v2) for all v1+v2 = n+ 1 +k when k = 0 and k = 1. We will prove this for any k ∈ {2, . . . , n} base on the results for k = 0 and k = 1. If the auction get the revenue n+ 2 for the profile (v1, n + 1−v1, k), it must be f1(n+ 1−v1, k) =v1 and f2(v1, k) = n+ 1−v1 since f3(v1, v2) = 1. Similarly, since f3(v1, n+2−v1) = 1, if the auction get the revenuen+2 for the profile (v1, n+2−v1, k), it must bef1(n+ 2−v1, k)≤v1 andf2(v1, k)≤n+ 2−v1. By the construction of layer kand the property of exact cover, for any (v1, v2) such thatv1+v2=n+ 1 +k, either Pr[(v1, n+ 1−v1, k)] =nand Pr[(n+ 2−v2, v2, k)] =n or Pr[(v1, n+ 2−v1, k)] =n and Pr[(n+ 1−v2, v2, k)] =n. So the revenue from bidder 1 and bidder 2 is at most n+ 1−v1+n+ 2−v2 =n+ 2−k. To achieve the target revenue n+ 2, the price for bidder 3 must be k. Therefore, we show thatf3(v1, v2) =h(v1, v2) for all (v1, v2) such thatv1+v2 ≥n+ 1.
Combining the construction for v3 = n and v3 < n, we can easily show that the graph Ghas a vertex cover with size at most k iff there exists a truthful auction can
get the probabilitym·n+n−k+s·n where s is the total number of profiles with positive probabilities whenv3 < n.
We are able to show a similar NP-completeness result for risk-averse sellers
Theorem 4.10. It is NP-complete to compute the optimal utility auction to maximize a concave function of the revenue for three correlated bidders.
Proof. Its membership in NP is obvious by similar arguments in the proof of The- orem 4.9. We prove NP-hardness by reducing from the following 3D-point-cover
problem.
Problem 1 (Problem 6.3 in [49]). Given a set S of n points (xi, yi, zi)(i = 1, ..., n)
and an integerk, recognize whether there existklines parallel to the axes whose union containsS.
Given an instance of 3D-point-cover problem in [2..n]3, we construct the fol-
lowing valuation distribution and concave utility function. u(x) = x if x ≤ 4 and u(x) = 4 otherwise. Pr[(1,1,1)] = n9, Pr[(x,1,1)] = Pr[(1, y,1)] = Pr[(1,1, z)] = n7 and Pr[(1, y, z)] = Pr[(x,1, z)] = Pr[(x, y,1)] = 1. In addition, Pr[(x, y, z)] = n3 if (x, y, z) ∈ S and 0 otherwise. It is easy to see that the optimal auction will set f1(1,1) =f2(1,1) =f3(1,1) = 1 to get utility 3 from (1,1,1) and f1(1, z) =f1(y,1) = f2(x,1) =f2(1, z) = f3(x,1) =f3(1, y) = 1 to get utility 3 from (x,1,1), (1, y,1) and (1,1, z) for any x, y, z≥2.
We will show that the optimal utility for this instance is at least 3n9+ 3n7(n− 1) + 4n3· |S|+ 9(n−1)2−kiff there exists akcover for the corresponding
3D-point- coverinstance. For sufficiency, suppose there exists a kcoverK for the instance. We
construct the following auction where f1(x, y) = 2 if the line of (x, y) is in the cover, andf1(x, y) = 1 otherwise. It is easy to check for any (x, y, z) with positive probability, at least one of (x, y), (y, z), (x, z) is 2, so the revenue is at least 4. So the total utility is 3n9+ 3n7(n−1) + 4n3· |S|+ 9(n−1)2−k. For necessity, if there exists an auction has utility 3n9+ 3n7(n−1) + 4n3· |S|+ 9(n−1)2−k, letK be the set of allf1(y, z)>1. It is easy to check this is a point cover for the instance.